We know that,
two angles which are formed by the intersection of two lines ,are <u>equal</u> and are called <u>vertically</u><u> </u><u>opposite</u><u> </u><u>angles.</u>
<u>ATQ,</u>
angle x lies on a line same as the other given angle which is intersected by an another line ,
hence,
From the statement of the problem, we have:
• a right triangle △ABC,
,• the altitude to the hypotenuse is denoted AN,
,• AB = 2√5 in,
,• NC = 1 in.
Using the data above, we draw the following diagram:
We must compute BN, AN and AC.
To solve this problem, we will use Pitagoras Theorem, which states that:
Where h is the hypotenuse, a and b the sides of a right triangle.
(I) From the picture, we see that we have two sub right triangles:
1) △ANC with sides:
• h = AC,
,• a = ,NC = 1,,
,• b = NA.
2) △ANB with sides:
• h = ,AB = 2√5,,
,• a = BN,
,• b = NA,
Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:
Equalling the last two equations, we have:
(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:
3) △ABC has sides:
• h = BC = ,BN + 1,,
,• a = AC,
,• b = ,AB = 2√5,,
Replacing these data in Pitagoras Theorem, we have:
Equalling the last equation to the one from (I), we have:
(III) Solving for BN the last quadratic equation, we get two values:
Because BN is a length, we must discard the negative value. So we have:
Replacing this value in the equation for AC, we get:
Finally, replacing the value of AC in the equation of NA, we get:
Answers
The lengths of the sides are:
• BN = 4 in,
,• AN = 2 in,
,• AC = √5 in.
